Why is it, in python/numpy:
from numpy import asarray
bools=asarray([False,True])
print(bools)
[False True]
print(1*bools, 0+bools, 0-bools) # False, True are valued as 0, 1
[0 1] [0 1] [ 0 -1]
print(-2*bools, -bools*2) # !? expected same result! :-/
[0 -2] [2 0]
print(-bools) # this is the reason!
[True False]
I consider it weird that -bools returns logical_not(bools), because in all other cases the behaviour is "arithmetic", not "logical".
One who wants to use an array of booleans as a 0/1 mask (or "characteristic function") is forced to use somehow involute expressions such as (0-bools) or (-1)*bools, and can easily incur into bugs if he forgets about this.
Why is it so, and what would be the best acceptable way to obtain the desired behaviour? (beside commenting of course)
Its all about operator order and data types.
>>> import numpy as np
>>> B = np.array([0, 1], dtype=np.bool)
>>> B
array([False, True], dtype=bool)
With numpy, boolean arrays are treated as that, boolean arrays. Every operation applied to them, will first try to maintain the data type. That is way:
>>> -B
array([ True, False], dtype=bool)
and
>>> ~B
array([ True, False], dtype=bool)
which are equivalent, return the element-wise negation of its elements. Note however that using -B throws a warning, as the function is deprecated.
When you use things like:
>>> B + 1
array([1, 2])
B and 1 are first casted under the hood to the same data type. In data-type promotions, the boolean array is always casted to a numeric array. In the above case, B is casted to int, which is similar as:
>>> B.astype(int) + 1
array([1, 2])
In your example:
>>> -B * 2
array([2, 0])
First the array B is negated by the operator - and then multiplied by 2. The desired behaviour can be adopted either by explicit data conversion, or adding brackets to ensure proper operation order:
>>> -(B * 2)
array([ 0, -2])
or
>>> -B.astype(int) * 2
array([ 0, -2])
Note that B.astype(int) can be replaced without data-copy by B.view(np.int8), as boolean are represented by characters and have thus 8 bits, the data can be viewed as integer with the .view method without needing to convert it.
>>> B.view(np.int8)
array([0, 1], dtype=int8)
So, in short, B.view(np.int8) or B.astype(yourtype) will always ensurs that B is a [0,1] numeric array.
Numpy arrays are homogenous—all elements have the same type for a given array, and the array object stores what type that is. When you create an array with True and False, it is an array of type bool and operators behave on the array as such. It's not surprising, then, that you get logical negation happening in situations that would be logical negation for a normal bool. When you use the arrays for integer math, then they are converted to 1's and 0's. Of all your examples, those are the more anomalous cases, that is, it's behavior that shouldn't be relied upon in good code.
As suggested in the comments, if you want to do math with an array of 0's and 1's, it's better to just make an array of 0's and 1's. However, depending on what you want to do with them, you might be better served looking into functions like numpy.where().
Related
Consider some array arr and advanced indexing mask mask:
import numpy as np
arr = np.arange(4).reshape(2, 2)
mask = A < 2
Using advanced indexing creates a new copy of an array. Accordingly, one cannot "chain" a mask with an an additional mask or even with a basic slicing operation to replace elements of an array:
submask = [False, True]
arr[mask][submask] = -1 # chaining 2 masks
arr[mask][:] = -1 # chaining a mask with a basic slicing operation
print(arr)
[[0 1]
[2 3]]
I have two related questions:
1/ What is the best way to replace elements of an array using chained masks?
2/ If advanced indexing returns a copy of an array, why does the following work?
arr[mask] = -1
print(arr)
[[-1 -1]
[ 2 3]]
The short answer:
you have to figure out a way of combining the masks. Since masks can "chain" in different ways I don't think there's a simple all-purpose substitute.
indexing can either be a __getitem__ call, or a __setitem__. Your last case is a set.
With chained indexing, a[mask1][mask2] =value gets translated into
a.__getitem__(mask1).__setitem__(mask2, value)
Whether a gets modified or not depends on what the first getitem produces (a view vs copy).
In [11]: arr = np.arange(4).reshape(2,2)
In [12]: mask = arr<2
In [13]: mask
Out[13]:
array([[ True, True],
[False, False]])
In [14]: arr[mask]
Out[14]: array([0, 1])
Indexing with a list or array may preserve the number of dimensions, but a boolean like this returns a 1d array, the items where the mask is true.
In your example, we could tweak the mask (details may vary with the intent of the 2nd mask):
In [15]: mask[:,0]=False
In [16]: mask
Out[16]:
array([[False, True],
[False, False]])
In [17]: arr[mask]
Out[17]: array([1])
In [18]: arr[mask] += 10
In [19]: arr
Out[19]:
array([[ 0, 11],
[ 2, 3]])
Or a logical combination of masks:
In [26]: (np.arange(4).reshape(2,2)<2)&[False,True]
Out[26]:
array([[False, True],
[False, False]])
Couple of good questions! My take:
I would do something like this:
x,y=np.where(mask)
arr[x[submask],y[submask]] = -1
From the official document:
Most of the following examples show the use of indexing when referencing data in an array. The examples work just as well when assigning to an array. See the section at the end for specific examples and explanations on how assignments work.
which means arr[mask]=1 is referrencing, while arr[mask] is extracting data and creates a copy.
I find that attempting to perform multiple boolean comparisons on numpy ndarrays using &, |, ==, >=, etc. often gives unexpected results, where the pure python order of operations seems on the surface to be violated (I was wrong about this; for example, True | False==True yields True). What are the "rules" or things going on under the hood that explain these results? Here are a few examples:
Comparing a boolean ndarray to the results of an elementwise comparison on a non-boolean ndarray:
In [36]: a = np.array([1,2,3])
In [37]: b = np.array([False, True, False])
In [38]: b & a==2 # unexpected, with no error raised!
Out[38]: array([False, False, False], dtype=bool)
In [39]: b & (a==2) # enclosing in parentheses resolves this
Out[39]: array([False, True, False], dtype=bool)
Elementwise &/| on boolean and non-boolean ndarrays:
In [79]: b = np.array([True,False,True])
In [80]: b & a # comparison is made, then array is re-cast into integers!
Out[80]: array([1, 0, 1])
Finding elements of array within two values:
In [47]: a>=2 & a<=2 # have seen this in different stackexchange threads
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
In [48]: (a>=2) & a<=2 # similar to behavior in In[38], but instead get *True* boolean array of
Out[48]: array([ True, True, True], dtype=bool)
In [49]: (a>=2) & (a<=2) # expected results
Out[49]: array([False, True, False], dtype=bool)
Logical &/| yielding results not in or [0,1] (which would be expected if a boolean result was coerced back into int).
In [90]: a & 2
Out[90]: array([0, 2, 2])
I welcome additional examples of this behavior.
I think you are confused about the precedence of the & | binary operators vs the comparison operators:
>>> import dis
>>> dis.dis("b & a==2")
1 0 LOAD_NAME 0 (b)
2 LOAD_NAME 1 (a)
4 BINARY_AND
6 LOAD_CONST 0 (2)
8 COMPARE_OP 2 (==)
10 RETURN_VALUE
You can see here that BINARY_AND is done first (between b and a) then the result is compared against 2 which, since it is a boolean array, is all False
The reason & and | have lower precedence is because they are not intended as logical operators, it represents the binary (math?) operation which numpy happens to use for logic, for example with ints I'd definitely expect the & to happen first:
if 13 & 7 == 5:
It is unfortunate that numpy cannot override the behaviour of the logical and and or operators since their precedence makes sense as logical operators but unfortunately they cannot be overridden so we just have to live will adding lots of brackets when doing boolean arrays.
Note that there was a proposal to allow and or to be overloaded but was not passed since basically it would only be a small convinience for numpy while making all other strict boolean operations slower.
a>=2 & a<=2 is evaluated as a>=(2 & a)<=2
The () part evaluates to array([0, 0, 2], dtype=int32)
a>=(2 & a) is a boolean array. But it is part of a Python a<x<b expression, which internally uses short circuiting. That is, it evaluates a<x and depending its value might actually skip the <b part. Something like True if a<x else x<b.
The familiar ValueError ambiguous arises when a boolean array is used in a scalar Python boolean context.
I've read through most of How to convert a boolean array to an int array , but I was still at loss with, how to (most efficiently) convert a numpy bool array to an int array, but with distinct values. For instance, I have:
>>> k=np.array([True, False, False, True, False])
>>> print k
[ True False False True False]
I'd like this to be converted to an array, where True is say, 2, and False is say 5.
Of course, I can always set up a linear equation:
>>> print 5-k*3
[2 5 5 2 5]
... which, while vectorized, needlessly employs both addition (subtraction) and multiplication; furthermore, if I'd want the opposite values (5 for True and 2 for False), I basically have to use (and recalculate) a different equation:
>>> print 2+k*3
[5 2 2 5 2]
... which is a bit of a readability issue for me.
In essence, this is merely a selection/mapping operation - but one which I'd like done in the numpy domain. How could I do that?
Seems like numpy.where is exactly what you want:
>>> import numpy as np
>>> k = np.array([True, False, False, True, False])
>>> np.where(k, 2, 5)
array([2, 5, 5, 2, 5])
Well, it seems I have to make a np.array (not a Python list!) containing the two distinct values:
>>> z=np.array([2,5])
>>> print z
[2 5]
... and then I can simply cast the boolean array (k) to int, and use that as selection indices in the "distinct values array" (z):
>>> print z[k.astype(int)]
[5 2 2 5 2]
It is also trivial to change the [2,5] to [5,2], so that is good.
I'm just not sure if this is the right way to do it (that is, maybe there exists something like (pseudocode) k.asdistinctvalues([2,5]) or something like that?)
In Numpy, nonzero(a), where(a) and argwhere(a), with a being a numpy array, all seem to return the non-zero indices of the array. What are the differences between these three calls?
On argwhere the documentation says:
np.argwhere(a) is the same as np.transpose(np.nonzero(a)).
Why have a whole function that just transposes the output of nonzero ? When would that be so useful that it deserves a separate function?
What about the difference between where(a) and nonzero(a)? Wouldn't they return the exact same result?
nonzero and argwhere both give you information about where in the array the elements are True. where works the same as nonzero in the form you have posted, but it has a second form:
np.where(mask,a,b)
which can be roughly thought of as a numpy "ufunc" version of the conditional expression:
a[i] if mask[i] else b[i]
(with appropriate broadcasting of a and b).
As far as having both nonzero and argwhere, they're conceptually different. nonzero is structured to return an object which can be used for indexing. This can be lighter-weight than creating an entire boolean mask if the 0's are sparse:
mask = a == 0 # entire array of bools
mask = np.nonzero(a)
Now you can use that mask to index other arrays, etc. However, as it is, it's not very nice conceptually to figure out which indices correspond to 0 elements. That's where argwhere comes in.
I can't comment on the usefulness of having a separate convenience function that transposes the result of another, but I can comment on where vs nonzero. In it's simplest use case, where is indeed the same as nonzero.
>>> np.where(np.array([[0,4],[4,0]]))
(array([0, 1]), array([1, 0]))
>>> np.nonzero(np.array([[0,4],[4,0]]))
(array([0, 1]), array([1, 0]))
or
>>> a = np.array([[1, 2],[3, 4]])
>>> np.where(a == 3)
(array([1, 0]),)
>>> np.nonzero(a == 3)
(array([1, 0]),)
where is different from nonzero in the case when you wish to pick elements of from array a if some condition is True and from array b when that condition is False.
>>> a = np.array([[6, 4],[0, -3]])
>>> b = np.array([[100, 200], [300, 400]])
>>> np.where(a > 0, a, b)
array([[6, 4], [300, 400]])
Again, I can't explain why they added the nonzero functionality to where, but this at least explains how the two are different.
EDIT: Fixed the first example... my logic was incorrect previously
I want to calculate an indexed weight sum across a large (1,000,000 x
3,000) boolean numpy array. The large boolean array changes
infrequently, but the weights come at query time, and I need answers
very fast, without copying the whole large array, or expanding the
small weight array to the size of the large array.
The result should be an array with 1,000,000 entries, each having the
sum of the weights array entries corresponding to that row's True
values.
I looked into using masked arrays, but they seem to require building a
weights array the size of my large boolean array.
The code below gives the correct results, but I can't afford that copy
during the multiply step. The multiply isn't even necessary, since
the values array is boolean, but at least it handles the broadcasting
properly.
I'm new to numpy, and loving it, but I'm about to give up on it for
this particular problem. I've learned enough numpy to know to stay
away from anything that loops in python.
My next step will be to write this routine in C (which has the added
benefit of letting me save memory by using bits instead of bytes, by
the way.)
Unless one of you numpy gurus can save me from cython?
from numpy import array, multiply, sum
# Construct an example values array, alternating True and False.
# This represents four records of three attributes each:
# array([[False, True, False],
# [ True, False, True],
# [False, True, False],
# [ True, False, True]], dtype=bool)
values = array([(x % 2) for x in range(12)], dtype=bool).reshape((4,3))
# Construct example weights, one for each attribute:
# array([1, 2, 3])
weights = array(range(1, 4))
# Create expensive NEW array with the weights for the True attributes.
# Broadcast the weights array into the values array.
# array([[0, 2, 0],
# [1, 0, 3],
# [0, 2, 0],
# [1, 0, 3]])
weighted = multiply(values, weights)
# Add up the weights:
# array([2, 4, 2, 4])
answers = sum(weighted, axis=1)
print answers
# Rejected masked_array solution is too expensive (and oddly inverts
# the results):
masked = numpy.ma.array([[1,2,3]] * 4, mask=values)
The dot product (or inner product) is what you want. It allows you to take a matrix of size m×n and a vector of length n and multiply them together yielding a vector of length m, where each entry is the weighted sum of a row of the matrix with the entries of the vector of as weights.
Numpy implements this as array1.dot(array2) (or numpy.dot(array1, array2) in older versions). e.g.:
from numpy import array
values = array([(x % 2) for x in range(12)], dtype=bool).reshape((4,3))
weights = array(range(1, 4))
answers = values.dot(weights)
print answers
# output: [ 2 4 2 4 ]
(You should benchmark this though, using the timeit module.)
It seems likely that dbaupp's answer is the correct one. But just for the sake of diversity, here's another solution that saves memory. This will work even for operations that don't have a built-in numpy equivalent.
>>> values = numpy.array([(x % 2) for x in range(12)], dtype=bool).reshape((4,3))
>>> weights = numpy.array(range(1, 4))
>>> weights_stretched = numpy.lib.stride_tricks.as_strided(weights, (4, 3), (0, 8))
numpy.lib.stride_tricks.as_strided is a wonderful little function! It allows you to specify shape and strides values that allow a small array to mimic a much larger array. Observe -- there aren't really four rows here; it just looks that way:
>>> weights_stretched[0][0] = 4
>>> weights_stretched
array([[4, 2, 3],
[4, 2, 3],
[4, 2, 3],
[4, 2, 3]])
So instead of passing a huge array to MaskedArray, you can pass a smaller one. (But as you've already noticed, numpy masking works in the opposite way you might expect; truth masks, rather than revealing, so you'll have to store your values inverted.) As you can see, MaskedArray doesn't copy any data; it just reflects whatever is in weights_stretched:
>>> masked = numpy.ma.MaskedArray(weights_stretched, numpy.logical_not(values))
>>> weights_stretched[0][0] = 1
>>> masked
masked_array(data =
[[-- 2 --]
[1 -- 3]
[-- 2 --]
[1 -- 3]],
mask =
[[ True False True]
[False True False]
[ True False True]
[False True False]],
fill_value=999999)
Now we can just pass it to sum:
>>> sum(masked, axis=1)
masked_array(data = [2 4 2 4],
mask = [False False False False],
fill_value=999999)
I benchmarked numpy.dot and the above against a 1,000,000 x 30 array. This is the result on a relatively modern MacBook Pro (numpy.dot is dot1; mine is dot2):
>>> %timeit dot1(values, weights)
1 loops, best of 3: 194 ms per loop
>>> %timeit dot2(values, weights)
1 loops, best of 3: 459 ms per loop
As you can see, the built-in numpy solution is faster. But stride_tricks is worth knowing about regardless, so I'm leaving this.
Would this work for you?
a = np.array([sum(row * weights) for row in values])
This uses sum() to immediately sum the row * weights values, so you don't need the memory to store all the intermediate values. Then the list comprehension collects all the values.
You said you want to avoid anything that "loops in Python". This at least does the looping with the C guts of Python, rather than an explicit Python loop, but it can't be as fast as a NumPy solution because that uses compiled C or Fortran.
I don't think you need numpy for something like that. And 1000000 by 3000 is a huge array; this will not fit in your RAM, most likely.
I would do it this way:
Let's say that you data is originally in a text file:
False,True,False
True,False,True
False,True,False
True,False,True
My code:
weight = range(1,4)
dicto = {'True':1, 'False':0}
with open ('my_data.txt') as fin:
a = sum(sum(dicto[ele]*w for ele,w in zip(line.strip().split(','),weight)) for line in fin)
Result:
>>> a
12
EDIT:
I think I slightly misread the question first time around, and summed up the everything together. Here is the solution that gives the exact solution that OP is after:
weight = range(1,4)
dicto = {'True':1, 'False':0}
with open ('my_data.txt') as fin:
a = [sum(dicto[ele]*w for ele,w in zip(line.strip().split(','),weight)) for line in fin]
Result:
>>> a
[2, 4, 2, 4]